Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <cstdio>

int n, *p, *q, front, ans;

int main() {
	scanf("%d", &n);
	p = new int[n];
	q = new int[n];
	for (int i = 0; i < n; ++i) {
		scanf("%d", p + i);
		q[--p[i]] = i;
	}
	for (int i = n - 2; i >= 0; --i) {
		if(q[i] > q[i + 1]) {
			front = i + 1;
			ans = 1;
			while(i--) {
				if(!(q[front] < q[i] && q[i] < q[i + 1]) && !(q[i] < q[i + 1] && q[i + 1] < q[front]) && !(q[i + 1] < q[front] && q[front] < q[i])) {
					++ans;
					front = i + 1;
				}
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}